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NCERT Solutions for Class 6 Maths Chapter 2 Whole Numbers

NCERT solutions for class 6 maths chapter 2 Whole Numbers Topic: Predecessor and Successor

Q1 Write the predecessor and successor of 19; 1997; 12000; 49; 100000.

Answer: The predecessor and successor of:

  • 19:

    • Predecessor : 18

    • Successor : 20

  • 1997:

    • Predecessor : 1996

    • Successor : 1998

  • 12000

    • Predecessor : 11999

    • Successor : 12001

  • 49:

    • Predecessor : 48

    • Successor : 50

  • 100000:

    • Predecessor : 99999

    • Successor : 100001

Q2 Is there any natural number that has no predecessor?

Answer: Every natural number has a predecessor. Although, it is interesting to know that the predecessor of 1 is not a natural number.

Q3 Is there any natural number which has no successor? Is there a last natural number?

Answer: Every natural number has a successor. There is no last natural number. There are infinite natural numbers.

NCERT solutions for class 6 maths chapter 2 Whole Numbers Topic: Whole Numbers

Q1 Are all natural numbers also whole numbers?

Answer: Yes, all the natural numbers are whole numbers. But, all whole numbers are not natural numbers.
Natural numbers = 1, 2, 3, 4, ....

Whole numbers= 1, 2, 3, 4, ....

Q2 Are all whole numbers also natural numbers?

Answer: No, all whole numbers are not natural numbers. 0 is a whole number, but it is not a natural number.

Q3 Which is the greatest whole number?

Answer: There are infinite whole numbers. Hence, there is no greatest whole number. Every whole number you can think of has a successor, which is greater than than the number.

NCERT solutions for class 6 maths chapter 2 Whole Numbers Exercise: 2.1

Q1 Write the next three natural numbers after 10999.

Answer: Given, 10999
The next three natural numbers are:
10999 + 1 = 11000
10999 + 2 = 11001
10999 + 3 = 11002

Q2 Write the three whole numbers occurring just before 10001.

Answer: Given, 10001
Three whole numbers occurring just before are:
11001 - 1 = 10000
11001 - 2 = 9999
11001 - 3 = 9998

Q3 Which is the smallest whole number?

Answer: The smallest whole number is 0. It has no whole number predecessor.

Q4 How many whole numbers are there between 32 and 53?

Answer: Given numbers are: 32 and 53

Number of whole numbers between 32 and 53 = (53-32) - 1 = 21 - 1 = 20

There are 20 whole numbers between 32 and 53

Q5 Write the successor of :
(a) 2440701 (b) 100199 (c) 1099999 (d) 2345670

Answer: The successor of following numbers are:

  • (a) 2440701

    • 2440701+1 = 2440702

  • (b) 100199

    • 100199 +1 = 100200

  • (c) 1099999

    • 1099999+1 = 1100000

  • (d) 2345670

    • 2345670+1 = 2345671

Q6 Write the predecessor of :
(a) 94 (b) 10000 (c) 208090 (d) 7654321

Answer: The predecessor of the following numbers are:

  • (a) 94

    • 94-1=93

  • (b) 10000

    • 10000 -1=9999

  • (c) 208090

    • 208090 -1=208089

  • (d) 7654321

    • 7654321-1=7654320

Q7 In each of the following pairs of numbers, state which whole number is on the left of the other number on the number line. Also write them with the appropriate sign (>, <) between them.
(a) 530, 503 (b) 370, 307 (c) 98765, 56789 (d) 9830415, 10023001

Answer: The number on the left on the number line is smaller than the number that is on the right on the number line.

  • (a) 530, 503

    • 503 is on the left.

    • \therefore 530 > 503

  • (b) 370, 307

    • 307 is on the left.

    • \therefore 370> 307

  • (c) 98765, 56789

    • 56789 is on the left.

    • \therefore 98765> 56789

  • (d) 9830415, 10023001

    • 9830415 is on the left.

    • \therefore 9830415< 10023001

Q8 Which of the following statements are true (T) and which are false (F)?
(a) Zero is the smallest natural number.
(b) 400 is the predecessor of 399.
(c) Zero is the smallest whole number.
(d) 600 is the successor of 599.
(e) All natural numbers are whole numbers.
(f) All whole numbers are natural numbers.
(g) The predecessor of a two-digit number is never a single-digit number.
(h) 1 is the smallest whole number.
(i) The natural number 1 has no predecessor.
(j) The whole number 1 has no predecessor.
(k) The whole number 13 lies between 11 and 12.
(l) The whole number 0 has no predecessor.
(m) The successor of a two-digit number is always a two-digit number

Answer:

(a) Zero is the smallest natural number. - False. 0 is not a natural number.
(b) 400 is the predecessor of 399. - False. 400 is the successor of 399.
(c) Zero is the smallest whole number. - True.
(d) 600 is the successor of 599. - True
(e) All natural numbers are the whole numbers.- True.
(f) All whole numbers are natural numbers.- False. 0 is a whole number but not a natural number.
(g) The predecessor of a two-digit number is never a single-digit number.- False. The predecessor of 10 is 9.
(h) 1 is the smallest whole number. - False. 0 is the smallest whole number.
(i) The natural number 1 has no predecessor. - True.
(j) The whole number 1 has no predecessor. - False. The whole number 1 has 0 as its predecessor.
(k) The whole number 13 lies between 11 and 12.- False. The whole number 13 lies on the right side of 12 on the number line.
(l) The whole number 0 has no predecessor.- True.
(m) The successor of a two-digit number is always a two-digit number- False. The successor of 99 is 100.

NCERT solutions for class 6 maths chapter 2 Whole Numbers Topic: Properties of Whole Numbers

Q Find : 7 + 18 + 13; 16 + 12 + 4.

Answer: 7 + 18 + 13; 16 + 12 + 4

NCERT solutions for class 6 maths chapter 2 Whole Numbers Exercise: 2.2

Q1 Find the sum by suitable rearrangement:
(a) 837 + 208 + 363 (b) 1962 + 453 + 1538 + 647

Answer: Sum by suitable rearrangement:

  • (a) 837 + 208 + 363
    837 + 208 + 363 =837 + 363+ 208
    = (837 + 363)+ 208
    = 1200 + 208
    = 1408

  • (b) 1962 + 453 + 1538 + 647
    1962 + 453 + 1538 + 647 = 1962+ 1538 + 453 + 647
    = (1962+ 1538) + (453 + 647)
    = 3500 + 1100
    = 4600

Q2 Find the product by suitable rearrangement:
(a) 2 × 1768 × 50 (b) 4 × 166 × 25
(c) 8 × 291 × 125 (d) 625 × 279 × 16
(e) 285 × 5 × 60 (f) 125 × 40 × 8 × 25

Answer: The product of the following by suitable rearrangement are:

  • (a) 2 \times 1768 \times 50
    \\ = 2 \times 50\times 1768 \\ = (2 \times 50)\times 1768 \\ = 100\times 1768 \\ = 176800

  • (b) 4 \times 166 \times 25
    \\ = 4\times 25 \times 166 \\ = (4\times 25 )\times 166 \\ = 100 \times 166 \\ = 16600

  • (c) 8 \times 291 \times125
    \\ = 8 \times125 \times 291 \\ =( 8 \times125) \times 291 \\ = 1000\times 291 \\ = 291000

  • (d) 625 \times 279 \times 16
    \\ = 625 \times 16\times 279 \\ = (625 \times 16)\times 279 \\ = 10000\times 279 \\ = 2790000

  • (e) 285 \times 5 \times 60
    \\ = 285 \times (5 \times 60) \\ = 285 \times 300 \\ = 85500

  • (f) 125 \times 40 \times 8 \times 25
    \\ = 125\times 8 \times 40 \times 25 \\ = (125\times 8 )\times (40 \times 25) \\ = 1000 \times 1000 \\ = 1000000

Q3 Find the value of the following:
(a) 297 × 17 + 297 × 3 (b) 54279 × 92 + 8 × 54279
(c) 81265 × 169 – 81265 × 69 (d) 3845 × 5 × 782 + 769 × 25 × 218

Answer:

  • (a) 297 \times 17 + 297 \times 3
    Using Distributive law.
    \\ = 297 \times (17 +3) \\ = 297 \times 20 \\ = 5940

  • (b) 54279 \times 92 + 8 \times 54279
    Using Commutative under multiplication
    54279 \times 92 + 54279 \times 8
    Using Distributive law.
    \\ = 54279 \times( 92 +8) \\ = 54279 \times 100 \\ = 5427900

  • (c) 81265 \times 169 - 81265 \times 69
    Using Distributive law.
    \\ = 81265 \times (169 - 69) \\ = 81265 \times 100 \\ = 8126500

  • (d) 3845 \times 5 \times 782 + 769 \times 25 \times 218
    \\ = (3845 \times 5) \times 782 + (769 \times 25) \times 218 \\ = 19225 \times 782 + 19225\times 218
    Using distributive law.
    \\ = 19225 \times( 782 + 218) \\ = 19225 \times1000 \\ = 19225000

Q4 Find the product using suitable properties.
(a) 738 × 103 (b) 854 × 102
(c) 258 × 1008 (d) 1005 × 168

Answer: The product of the folllowing using suitable properties are:

  • (a) 738 \times 103
    \\ = 738 \times (100+3)
    Using distributive law.
    \\ = 738 \times 100+738 \times3 \\ = 73800+2214 \\ = 76014

  • (b) 854 \times 102
    \\ = 854 \times (100+2)
    Using distributive law.
    \\ = 854 \times 100+854 \times 2 \\ = 85400+1708 \\ = 87108

  • (c) 258 \times 1008
    \\ = 258 \times (1000+8)
    Using Distributive law.
    \\ = 258 \times 1000+258 \times8 \\ = 258000+2064 \\ = 260064

  • (d) 1005 \times 168
    \\ = (1000+5) \times 168
    Using Distributive law.
    \\ = 1000\times 168+5 \times 168 \\ = 168000+840 \\ = 168840

Q5 A taxidriver filled his car petrol tank with 40 litres of petrol on Monday. The next day, he filled the tank with 50 litres of petrol. If the petrol costs rupees 44 per litre, how much did he spend in all on petrol?

Answer: Amount of petrol filled on Monday = 40\ litres
Amount of petrol filled on Tuesday = 40\ litres
\therefore Total amount of petrol = (40+40)\ litres = 80\ litres
Cost of 1 litre of petrol = Rs.\ 44
\therefore Cost of 80\ litres of petrol = Rs.\ (44\times80)
= Rs.\ 3520

Q6 A vendor supplies 32 litres of milk to a hotel in the morning and 68 litres of milk in the evening. If the milk costs rupees 45 per litre, how much money is due to the vendor per day?

Answer: Amount of milk supplied in the morning = 32\ litres
Amount of milk supplied in the evening = 68\ litres
\therefore Total amount of petrol = (32+68)\ litres = 100\ litres
Cost of 1 litre of milk = Rs.\ 45
\therefore Cost of 100\ litres of milk = Rs.\ (45\times100)
= Rs.\ 4500

Q7 Match the following:
(i) 425 × 136 = 425 × (6 + 30 +100) (a) Commutativity under multiplication.
(ii) 2 × 49 × 50 = 2 × 50 × 49 (b) Commutativity under addition.
(iii) 80 + 2005 + 20 = 80 + 20 + 2005 (c) Distributivity of multiplication over addition.

Answer:

(i) 425 \times 136 = 425 \times (6 + 30 +100)

(c) Distributivity of multiplication over addition.

(ii) 2 \times 49 \times 50 = 2 \times 50 \times 49

(a) Commutativity under multiplication.

(iii) 80 + 2005 + 20 = 80 + 20 + 2005

(b) Commutativity under addition.

NCERT solutions for class 6 maths chapter 2 Whole Numbers Topic: Patterns in Whole Numbers

Q1 Which numbers can be shown only as a line?

Answer: 1,5,7,11,13 can be shown only as a line. They cannot be shown as a rectangle or square or triangle.

Q2 Which can be shown as squares?

Answer: 4 and 9 can be shown as squares.
4: 2 rows and 2 columns.
9: 3 rows and 3 columns

Q3 Which can be shown as rectangles?

Answer: 4, 6, 8,10, 12 can be shown as rectangles. (Note: We are not counting squares as rectangles here)

Q4 Write down the first seven numbers that can be arranged as triangles, e.g. 3, 6, ...

Answer: 3, 6, 10, 15, 21, 28, 36.

Q5 Some numbers can be shown by two rectangles, for example


Give at least five other such examples.

Answer: We can represent a number by two rectangles. for example 12 = 3 x 4 or 2 x 6

five other such examples are :

  • 24 = 12 x 2 or 24 = 6 x 4

  • 18 = 9 x 2 or 18 = 3 x 6

  • 15 = 15 x 1 or 15 = 3 x 5

  • 30 = 10 x 3 or 30 = 5 x 6

  • 40 = 10 x 4 or 40 = 5 x 8.

NCERT solutions for class 6 maths chapter 2 Whole Numbers Exercise: 2.3

Q1 Which of the following will not represent zero:
(a) 1 + 0 (b) 0 × 0 (c) 0/ 2 (d) (10-10)/2

Answer:

  • (a) 1 + 0
    It does not represent zero.

  • (b) 0 × 0
    It represents zero.

  • (c) \frac{0}{2}=0
    It represents zero.

  • (d) \frac{10-10}{2}=0
    It represents zero.

Q2 If the product of two whole numbers is zero, can we say that one or both of them will be zero? Justify through examples.

Answer: If the product of 2 whole numbers is zero, then one of them is definitely zero.
For example, 0 x 2 = 0 and 17 x 0 = 0
If the product of 2 whole numbers is zero, then both of them may be zero.
0 x 0 = 0
However, 2 x 3 = 6 (Since numbers to be multiplied are not equal to zero, the result of the product will also be non-zero.)

Q3 If the product of two whole numbers is 1, can we say that one or both of them will be 1? Justify through examples.

Answer: If the product of 2 numbers is 1, then both the numbers have to equal to 1.
For example, 1 x 1 = 1
However, 1 x 6 = 6
Clearly, the product of two whole numbers will be 1 in the situation when both numbers to be multiplied are 1.

Q4 Find using distributive property :
(a) 728 \times 101 (b) 5437 \times 1001 (c) 824 \times 25 (d) 4275 \times 125 (e) 504 \times 35

Answer:

  • (a) 728 \times 101= 728 \times (100 + 1)

    • = 728 \times 100 + 728 \times 1

    • = 72800 + 728

    • = 73528

  • (b) 5437 \times 1001 = 5437 \times (1000 + 1)

    • = 5437 \times 1000 + 5437 \times 1

    • = 5437000 + 5437

    • = 5442437

  • (c) 824 \times 25 (800 + 24) \times 25 = (800 + 25 - 1) 25

    • =800 \times 25+25 x 25-1 \times 25

    • = 20000 + 625 - 25

    • = 20000 + 600

    • = 20600

  • (d) 4275 \times 125 = (4000 + 200 + 100 - 25) \times 125

    • = 4000 \times 125 + 200 \times 125 + 100 \times 125 - 25 \times 125

    • = 500000 + 25000 + 12500 - 3125

    • = 534375

  • (e) 504 \times 35 = (500 + 4) \times 35

    • = 500 x 35 +4 \times 35

    • = 17500 + 140

    • = 17640

Q5 Study the pattern :
\times 8 + 1 = 9 1234 \times 8 + 4 = 9876
12 \times 8 + 2 = 98 12345 \times 8 + 5 = 98765
123 \times 8 + 3 = 987
Write the next two steps. Can you say how the pattern works?
(Hint: 12345 = 11111 + 1111 + 111 + 11 + 1).

Answer: 123456 \times 8 + 6 = 987648 + 6 = 987654
1234567 \times 8 + 7 = 9876536 + 7 = 9876543
Yes, the pattern works.
As 123456 = 111111 + 11111 + 1111 + 111 + 11 + 1,
123456 \times 8 = (111111 + 11111 + 1111 + 111 + 11 + 1) \times 8
= 111111 \times 8 + 11111 \times 8 + 1111 \times 8 + 111 \times 8 + 11 \times 8 + 1 \times 8
= 888888 + 88888 + 8888 + 888 + 88 + 8
= 987648
And,
123456 \times 8 + 6 = 987648 + 6 = 987648

Responsive Table
Chapter No. Chapter Name
Chapter 1 NCERT Solutions for class 6 maths chapter 1 Knowing Our Numbers
Chapter 2 NCERT solutions for class 6 maths chapter 2 Whole Numbers
Chapter 3 NCERT solutions for class 6 maths chapter 3 Playing with Numbers
Chapter 4 NCERT Solutions for class 6 maths chapter 4 Basic Geometrical Ideas
Chapter 5 NCERT solutions for class 6 maths chapter 5 Understanding Elementary Shapes
Chapter 6 NCERT solutions for class 6 maths chapter 6 Integers
Chapter 7 NCERT Solutions for class 6 maths chapter 7 Fractions
Chapter 8 NCERT solutions for class 6 maths chapter 8 Decimals
Chapter 9 NCERT solutions for class 6 maths chapter 9 Data Handling
Chapter 10 NCERT solutions for class 6 maths chapter 10 Mensuration
Chapter 11 NCERT Solutions for class 6 maths chapter 11 Algebra
Chapter 12 NCERT solutions for class 6 maths chapter 12 Ratio and Proportion
Chapter 13 NCERT solutions for class 6 maths chapter 13 Symmetry
Chapter 14 NCERT Solutions for class 6 maths chapter 14 Practical Geometry

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